## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

The free-fall acceleration on the newly discovered planet is $12.25~m/s^2$
We can write an expression for the free-fall acceleration on the earth. $g = \frac{G~M}{R^2} = 9.80~m/s^2$ We can find the free-fall acceleration $g'$ on the newly discovered planet. $g' = \frac{G~(5M)}{(2R)^2}$ $g' = (\frac{5}{4})~(\frac{G~M}{R^2})$ $g' = (\frac{5}{4})~(9.80~m/s^2)$ $g' = 12.25~m/s^2$ The free-fall acceleration on the newly discovered planet is $12.25~m/s^2$.